NDA Maths · Vectors

Vector Geometry — Triangles, Parallelograms, Quadrilaterals

Treating triangle / parallelogram / quadrilateral sides and diagonals as vectors, then using the loop identity, centroid formula and parallelogram law to extract distances and angles.

All Vectors notes NDA Maths strategy

Why this matters

Once vectors are anchored at an origin, plane figures (triangles, parallelograms, quadrilaterals) become vector equations you can solve algebraically. The closed-loop identity AB + BC + CA = 0 turns a triangle into one usable relation; the centroid is the average of the three vertex position vectors; parallelogram identities link sides to diagonals; named-vertex angles drop out of dot products on position vectors. The four concepts below are the levers — once you spot which one is in play, every PYQ resolves to a few lines of algebra. 11 PYQs across 2017–2026, mostly MODERATE.

Concept 1 of 4

Triangle closed-loop and centroid formula

Intuition

Walking around a triangle from

A

B

C

and back to

A

returns you to the starting point, so the three side-vectors taken in order must sum to the zero vector. The centroid is the average of the three vertex position vectors — geometrically the meeting point of the three medians, which it divides in a

2:1

ratio.

Definition

For a triangle $ABC$ , the side-vectors satisfy $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$ . The centroid $G$ has position vector $\vec{g} = \dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$ . Each median is divided by $G$ in ratio $2 : 1$ (longer part from vertex to $G$ ).

Loop identity + centroid

\vec{AB} + \vec{BC} + \vec{CA} = \vec{0} \qquad \vec{g} = \dfrac{\vec{a} + \vec{b} + \vec{c}}{3}

$\vec{a}, \vec{b}, \vec{c}$ position vectors of vertices $A, B, C$
$\vec{AB}$ side vector from $A$ to $B$ , equal to $\vec{b} - \vec{a}$
$\vec{g}$ position vector of the centroid $G$

Diagram · closed loop & centroid

Walking the edges A→B→C→A returns you to the start, so AB + BC + CA = 0. The three medians meet at the centroid G = (a + b + c)/3, the average of the vertices' position vectors.

Worked example

The vertices of a triangle have position vectors

\vec{a} = \hat{i} + 2\hat{j}

\vec{b} = 3\hat{i} + 4\hat{j}

\vec{c} = 5\hat{i} - \hat{j}

. Find

\vec{AG}

, where

G

is the centroid.

Compute the centroid: $\vec{g} = \dfrac{\vec{a} + \vec{b} + \vec{c}}{3} = \dfrac{(1+3+5)\hat{i} + (2+4-1)\hat{j}}{3} = 3\hat{i} + \dfrac{5}{3}\hat{j}$ .
Use $\vec{AG} = \vec{g} - \vec{a} = \left(3\hat{i} + \tfrac{5}{3}\hat{j}\right) - (\hat{i} + 2\hat{j}) = 2\hat{i} - \tfrac{1}{3}\hat{j}$ .

Answer:

\vec{AG} = 2\hat{i} - \dfrac{1}{3}\hat{j}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the centroid of the triangle with vertices

A(1, 2, 3)

B(3, -1, 0)

C(2, 2, 3)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\vec{AB} + \vec{BC} + \vec{CA} = ?$
2.
Centroid of $A(0,0)$ , $B(3,0)$ , $C(0,3)$ ?
3.
Centroid formula for vertices $\vec{a}, \vec{b}, \vec{c}$ ?
4.
The centroid divides each median in ratio?

From the bank · past-year question

Example 1VectorsEASY

\vec{a}, \vec{b}, \vec{c}

are the position vectors of the vertices

A, B, C

respectively of a triangle

ABC

and

G

is the centroid of the triangle, then what is

\overrightarrow{AG}

equal to?

[Q97 · Apr · 2022]

Direction matters in the loop — $\vec{BA} = -\vec{AB}$

The identity

\vec{AB}+\vec{BC}+\vec{CA}=\vec{0}

requires the sides to be traversed in one consistent direction around the triangle. If a statement reads

\vec{AB}+\vec{BC}-\vec{CA}=\vec{0}

, it's wrong — that's saying

\vec{AC}

instead of

\vec{CA}

, which reverses one side.

$\vec{AG}$ is NOT $\vec{g}/3$ — it is $\vec{g} - \vec{a}$

A common factor-of-3 distractor.

\vec{AG}

is the displacement from

A

G

, so it equals

\vec{g} - \vec{a}

. After algebra

\vec{AG} = \dfrac{(\vec{b}-\vec{a}) + (\vec{c}-\vec{a})}{3}

— two-thirds of the median from

A

Drill 1 more on triangle closed-loop and centroid formula

Concept 2 of 4

Parallelogram properties and diagonal relations

Intuition

A parallelogram has opposite sides equal and parallel as vectors. Its diagonals bisect each other, so the midpoint of one diagonal IS the midpoint of the other. Whenever a question mentions an origin

O

and a parallelogram, expect a clean linear identity built around that shared midpoint.

Definition

In parallelogram $ABCD$ (vertices in order): $\vec{AB} = \vec{DC}$ and $\vec{AD} = \vec{BC}$ . Diagonals: $\vec{AC} = \vec{AB} + \vec{AD}$ and $\vec{BD} = \vec{AD} - \vec{AB}$ , so $\vec{AB} = \dfrac{\vec{AC} - \vec{BD}}{2}$ and $\vec{AD} = \dfrac{\vec{AC} + \vec{BD}}{2}$ . From any origin $O$ : $\vec{OA} + \vec{OC} = \vec{OB} + \vec{OD} = 2\vec{OP}$ , where $P$ is the common midpoint of $AC$ and $BD$ .

Sides from diagonals

\vec{AB} = \tfrac{1}{2}(\vec{AC} - \vec{BD}), \quad \vec{AD} = \tfrac{1}{2}(\vec{AC} + \vec{BD}), \quad \vec{OA}+\vec{OC} = \vec{OB}+\vec{OD}

$ABCD$ parallelogram with vertices labelled in order
$\vec{AC}, \vec{BD}$ diagonal vectors
$O$ arbitrary origin (often the centre or an external point)

Diagram · parallelogram diagonals = a + b and a − b

From a shared corner, sides a and b span the parallelogram. The diagonal from that corner is a + b; the diagonal between the side tips is a − b. They bisect each other, and |a + b|² + |a − b|² = 2(|a|² + |b|²).

Worked example

In parallelogram

ABCD

(vertices in order), the position vectors of

A, B, C

are

\vec{a} = 2\hat{i} + \hat{j}

\vec{b} = 4\hat{i} + 3\hat{j}

\vec{c} = 6\hat{i} - \hat{j}

. Find the position vector of

D

In a parallelogram, $\vec{AD} = \vec{BC}$ , which gives $\vec{d} - \vec{a} = \vec{c} - \vec{b}$ .
So $\vec{d} = \vec{a} + \vec{c} - \vec{b} = (2+6-4)\hat{i} + (1-1-3)\hat{j}$ .
$= 4\hat{i} - 3\hat{j}$ .

Answer:

\vec{d} = 4\hat{i} - 3\hat{j}

Practice this conceptself-check · 4 quick reps

Try it yourself

In parallelogram

ABCD

\vec{AB} = 3\hat{i} + \hat{j}

and

\vec{AD} = \hat{i} + 2\hat{j}

. Find the diagonal

\vec{AC}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
In parallelogram $ABCD$ , $\vec{AB}$ equals which side?
2.
Diagonal $\vec{AC}$ in terms of the sides?
3.
Fourth vertex $D$ of parallelogram $ABCD$ ?
4.
Do the diagonals of a parallelogram bisect each other?

From the bank · past-year question

Example 2VectorsMODERATE

PQRS is a parallelogram. If

\overrightarrow{PR} = \vec{a}

and

\overrightarrow{QS} = \vec{b}

, then what is

\overrightarrow{PQ}

equal to?

[Q66 · Sep · 2022]

Vertex order $ABCD$ matters

If the vertices are listed in a non-cyclic order, the figure is NOT a parallelogram in the standard sense.

\vec{AB} = \vec{DC}

(not

\vec{CD}

) — the equal sides are the ones going in the SAME direction around the figure.

The fourth vertex of a parallelogram: $D = A + C - B$

A, B, C

are three consecutive vertices,

\vec{AD} = \vec{BC}

forces

\vec{d} = \vec{a} + \vec{c} - \vec{b}

. A factor-of-2 distractor here often offers

\vec{a} + \vec{c} - 2\vec{b}

; reject it.

Drill 2 more on parallelogram properties and diagonal relations

Concept 3 of 4

Angles and vertices from position vectors

Intuition

The angle at vertex

C

of a triangle is the angle between the two sides leaving

C

— namely

\vec{CA}

and

\vec{CB}

. Build those side-vectors from position vectors, then plug into the dot-product angle formula. Same idea works for the angle between diagonals of a quadrilateral.

Definition

If $A, B, C$ have position vectors $\vec{a}, \vec{b}, \vec{c}$ , the side vectors at $C$ are $\vec{CA} = \vec{a} - \vec{c}$ and $\vec{CB} = \vec{b} - \vec{c}$ . Then $\cos C = \dfrac{\vec{CA}\cdot\vec{CB}}{|\vec{CA}|\,|\vec{CB}|}$ . For a quadrilateral with diagonals $AC$ and $BD$ , the same formula applies with the two diagonal vectors.

Angle at vertex from position vectors

\cos C = \dfrac{(\vec{a} - \vec{c})\cdot(\vec{b} - \vec{c})}{|\vec{a} - \vec{c}|\,|\vec{b} - \vec{c}|}

$\vec{a}, \vec{b}, \vec{c}$ position vectors of the vertices
$C$ angle of the triangle at vertex $C$

Worked example

The vertices of triangle

ABC

have position vectors

\vec{a} = \hat{i} + \hat{j}

\vec{b} = 3\hat{i} + 5\hat{j}

\vec{c} = 5\hat{i} + \hat{j}

. Find the angle at vertex

C

Build the side vectors at $C$ : $\vec{CA} = \vec{a} - \vec{c} = -4\hat{i}$ , $\vec{CB} = \vec{b} - \vec{c} = -2\hat{i} + 4\hat{j}$ .
Dot product: $\vec{CA}\cdot\vec{CB} = (-4)(-2) + 0\cdot 4 = 8$ .
Magnitudes: $|\vec{CA}| = 4$ , $|\vec{CB}| = \sqrt{4 + 16} = 2\sqrt{5}$ .
Apply the angle formula: $\cos C = \dfrac{8}{4 \cdot 2\sqrt{5}} = \dfrac{1}{\sqrt{5}}$ .

Answer:

C = \cos^{-1}\!\left(\dfrac{1}{\sqrt{5}}\right)

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the angle between

\vec{p} = \hat{i} + \sqrt{3}\,\hat{j}

and

\vec{q} = \hat{i}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Side vector at vertex $C$ : $\vec{CA} = ?$
2.
$\cos C$ formula at vertex $C$ ?
3.
Angle between $\hat{i} + \hat{j}$ and $\hat{i}$ ?
4.
Angle between $\vec{a}$ and $-\vec{a}$ ?

From the bank · past-year question

Example 3VectorsMODERATE

The position vectors of vertices A, B and C of triangle ABC are respectively

\hat{j}+\hat{k}

3\hat{i}+\hat{j}+5\hat{k}

and

3\hat{j}+3\hat{k}

. What is angle C equal to?

[Q70 · Sep · 2022]

Direction of side vectors changes the angle

The angle at

C

is between

\vec{CA}

and

\vec{CB}

— NOT between

\vec{AC}

and

\vec{BC}

. Reversing both flips the dot-product sign and gives

\pi - C

instead of

C

. Always start from the named vertex outwards.

Fourth-vertex problems: $D = A + C - B$ , not the midpoint

If a parallelogram

ABCD

lists

A, B, C

as consecutive vertices, the fourth vertex

D

satisfies

\vec{AD} = \vec{BC}

, giving

\vec{d} = \vec{a} + \vec{c} - \vec{b}

. Mid-segment formulas applied here are the typical wrong-option trap.

Drill 3 more on angles and vertices from position vectors

Concept 4 of 4

Distance and perpendicularity identities in quadrilaterals

Intuition

Quadrilateral distance puzzles are almost always tests of vector arithmetic. Expand each squared distance as

|\vec{XY}|^2 = \vec{XY}\cdot\vec{XY}

, collect like terms, and watch dot products with shared edges cancel. The parallelogram law

|\vec{p}+\vec{q}|^2 + |\vec{p}-\vec{q}|^2 = 2(|\vec{p}|^2 + |\vec{q}|^2)

is the workhorse identity.

Definition

For any four points $P, Q, R, S$ with position vectors $\vec{p}, \vec{q}, \vec{r}, \vec{s}$ : $\vec{PQ}$ is parallel to $\vec{RS}$ iff $\vec{PQ} \times \vec{RS} = \vec{0}$ ; perpendicular iff $\vec{PQ} \cdot \vec{RS} = 0$ . Squared distances expand as $|\vec{PQ}|^2 = (\vec{q}-\vec{p})\cdot(\vec{q}-\vec{p}) = |\vec{q}|^2 - 2\vec{p}\cdot\vec{q} + |\vec{p}|^2$ .

Parallelogram law and distance expansion

|\vec{p}+\vec{q}|^2 + |\vec{p}-\vec{q}|^2 = 2(|\vec{p}|^2 + |\vec{q}|^2)

$\vec{p}, \vec{q}$ any two vectors (often diagonals or sides)
$|\vec{p}+\vec{q}|, |\vec{p}-\vec{q}|$ diagonal magnitudes when $\vec{p}, \vec{q}$ are sides

Worked example

Points have position vectors

\vec{p} = \hat{i}

\vec{q} = \hat{j}

\vec{r} = -\hat{i}

\vec{s} = -\hat{j}

. Check whether

\vec{PQ}

is parallel to

\vec{SR}

and whether

\vec{PR}

is perpendicular to

\vec{QS}

Compute the vectors: $\vec{PQ} = \vec{q} - \vec{p} = -\hat{i} + \hat{j}$ ; $\vec{SR} = \vec{r} - \vec{s} = -\hat{i} + \hat{j}$ .
They are equal (and hence parallel — in fact same direction): the figure has a pair of parallel sides.
Now $\vec{PR} = \vec{r} - \vec{p} = -2\hat{i}$ ; $\vec{QS} = \vec{s} - \vec{q} = -2\hat{j}$ .
Dot product: $\vec{PR}\cdot\vec{QS} = (-2)(0) + 0(-2) = 0$ , so the diagonals are perpendicular.

Answer:

\vec{PQ} \parallel \vec{SR}

and

\vec{PR} \perp \vec{QS}

Practice this conceptself-check · 4 quick reps

Try it yourself

For

\vec{p} = 2\hat{i} + \hat{j}

and

\vec{q} = \hat{i} - 3\hat{j}

, verify the parallelogram law

|\vec{p}+\vec{q}|^2 + |\vec{p}-\vec{q}|^2 = 2(|\vec{p}|^2 + |\vec{q}|^2)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Parallelogram law: $|\vec{a} + \vec{b}|^2 + |\vec{a} - \vec{b}|^2 = ?$
2.
$\vec{PQ} \perp \vec{RS}$ iff?
3.
Expand $|\vec{q} - \vec{p}|^2$ .
4.
$\vec{PQ} \parallel \vec{RS}$ iff $\vec{PQ}\times\vec{RS} = ?$

From the bank · past-year question

Example 4VectorsMODERATE

for the items that follow: P(-2,-3,5), Q(4,-1,5), R(6,-4,8) and S(2,-6,10) are four points.

Consider the following statements: I. PQ is parallel to RS. II. PR is perpendicular to QS. Which of the statements given above is/are correct?

[Q53 · Apr · 2026]

Direction of comparison matters for parallelism

Two vectors are parallel if one is a scalar multiple of the other — the scalar can be negative.

\vec{PQ}

and

\vec{RS}

point in opposite directions yet are still parallel. But if a PYQ asks whether

\vec{PQ}

and

\vec{SR}

are equal (not just parallel), the sign matters.

Expand squared distances algebraically — don't reach for coordinates first

An identity like

PQ^2 + 2QS^2 - 2PR^2 = ?

is much faster to verify by expanding each

|\cdots|^2

as a dot product and collecting terms in

\vec{p}\cdot\vec{q}

\vec{p}\cdot\vec{r}

, etc., than by plugging in coordinates and computing each squared distance separately.

Drill 1 more on distance and perpendicularity identities in quadrilaterals

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Triangle closed-loop and centroid formula
Loop identity + centroid
$\vec{AB} + \vec{BC} + \vec{CA} = \vec{0} \qquad \vec{g} = \dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$
Parallelogram properties and diagonal relations
Sides from diagonals
$\vec{AB} = \tfrac{1}{2}(\vec{AC} - \vec{BD}), \quad \vec{AD} = \tfrac{1}{2}(\vec{AC} + \vec{BD}), \quad \vec{OA}+\vec{OC} = \vec{OB}+\vec{OD}$
Angles and vertices from position vectors
Angle at vertex from position vectors
$\cos C = \dfrac{(\vec{a} - \vec{c})\cdot(\vec{b} - \vec{c})}{|\vec{a} - \vec{c}|\,|\vec{b} - \vec{c}|}$
Distance and perpendicularity identities in quadrilaterals
Parallelogram law and distance expansion
$|\vec{p}+\vec{q}|^2 + |\vec{p}-\vec{q}|^2 = 2(|\vec{p}|^2 + |\vec{q}|^2)$

Watch out for (8)

Direction matters in the loop — $\vec{BA} = -\vec{AB}$
→ Triangle closed-loop and centroid formula
$\vec{AG}$ is NOT $\vec{g}/3$ — it is $\vec{g} - \vec{a}$
→ Triangle closed-loop and centroid formula
Vertex order $ABCD$ matters
→ Parallelogram properties and diagonal relations
The fourth vertex of a parallelogram: $D = A + C - B$
→ Parallelogram properties and diagonal relations
Direction of side vectors changes the angle
→ Angles and vertices from position vectors
Fourth-vertex problems: $D = A + C - B$ , not the midpoint
→ Angles and vertices from position vectors
Direction of comparison matters for parallelism
→ Distance and perpendicularity identities in quadrilaterals
Expand squared distances algebraically — don't reach for coordinates first
→ Distance and perpendicularity identities in quadrilaterals

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1VectorsMODERATE

In a triangle ABC, if taken in order, consider the following statements: 1.

\vec{AB}+\vec{BC}+\vec{CA}=\vec{0}

\vec{AB}+\vec{BC}-\vec{CA}=\vec{0}

\vec{AB}-\vec{BC}+\vec{CA}=\vec{0}

\vec{BA}-\vec{BC}+\vec{CA}=\vec{0}

How many of the above statements are correct?

[Q70 · Sep · 2018]

Example 2VectorsEASY

Let

ABCD

be a parallelogram whose diagonals intersect at

P

and let

O

be the origin. What is

\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}

equal to?

[Q67 · Apr · 2017]

Example 3VectorsHARD

For the following two (02) items: Let

A(1,-1,0)

B(-2,1,8)

and

C(-1,2,7)

are three consecutive vertices of a parallelogram

ABCD

If angle

BCD

\theta

, then what is

\cos^2\theta

equal to?

[Q56 · Sep · 2025]

Example 4VectorsMODERATE

for the items that follow: P(-2,-3,5), Q(4,-1,5), R(6,-4,8) and S(2,-6,10) are four points.

What is

(PQ^2+2QS^2-2PR^2)

equal to?

[Q54 · Apr · 2026]

Example 5VectorsMODERATE

ABCD

is a quadrilateral whose diagonals are

AC

and

BD

. Which one of the following is correct?

[Q68 · Apr · 2017]

Drill every past-year question on this subtopic

11 questions from the bank — paginated, with cart and Word-export support.

Drill the 11 questions

Drill every past-year question on this subtopic

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